Write the proportions given by the similarity relationship and solve each equation for the numerators. Then, add them up and relate them with the perimeters of the triangles.
See solution.
Practice makes perfect
We are given that â–ł ABC ~ â–ł DEF and ABDE= mn. By definition of similar triangles, we can write the following proportions.
AB/DE=BC/EF=AC/DF = m/n
We can split the equality above into the following three equations.
AB = m/n* DE [0.2cm]
BC = m/n* EF [0.2cm]
AC = m/n* DF
Next, we write down the perimeter of each triangle, which is the sum of the lengths of its sides.
P_1 &= AB + BC + AC
P_2 &= DE + EF+ DF
Finally, we substitute the equations above into P_1.
We have obtained the desired conclusion, which is that the perimeters of similar triangles are proportional to the scale factor between the triangles.
Paragraph Proof
Given: & â–ł ABC ~ â–ł DEF, ABDE= mn
Prove: & perimeter ofâ–ł ABCperimeter ofâ–ł DEF = mn
Proof: Since â–ł ABC ~ â–ł DEF and ABDE= mn, by definition of similar triangles we get ABDE= BCEF= ACDF = mn. By solving each of these equations for the sides of â–ł ABC we obtain AB= mnDE, BC= mnEF, and AC= mnDF.
By adding the three equations before, we get on the left-hand side AB+BC+AC, which is the perimeter of â–ł ABC. On the right-hand side we get mn(DE+EF+DF). Notice that the expression between parentheses is the perimeter of â–ł DEF. Then, we get that P_(â–ł ABC) = mnP_(â–ł DEF), which implies that P_(â–ł ABC)P_(â–ł DEF) = mn.
